$x_1,x_2,x_3$  are positive roots of  $x^3-6x^2+3px-2p=0(p\in R)$  then the value of ${\text{sin}}^{-1}(\frac{1}{x_1}+\frac{1}{x_2})+{\cos }^{-1}(\frac{1}{x_2}+\frac{1}{x_3})-{\tan }^{-1}(\frac{1}{x_3}+\frac{1}{x_1})$  is equal to 
 

$x_1+x_2+x_3=6 x_1{x}_2+x_2{x}_{3+}{x}_3{x}_1=3p x_1{x}_2{x}_3=2p$

$\begin{array}{l}\frac{x_1+x_2+x_3}{3}\ge {(x_{1,}{x}_2,{\text{\hspace{0.17em}x}}_3\text{)}}^{1/3}\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}\le \text{4 }\mathrm{..............}\text{(1)}\\ x_1{x}_2{x}_3=2p\frac{3p}{3}\ge {\left[{\left(x_1{x}_2{x}_3\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{p}^3\ge 4p^{2}p\ge 4\mathrm{............}\text{(2)}\\ \text{From (1) and (2) p=4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{x}_1{x}_2+x_2{x}_{3+}{x}_3{x}_1=\text{ 12,} x_1{x}_2{x}_3=8,x_1=2,x_2=2,x_3=2\end{array}$${\text{sin}}^{-1}\left(1\right)+{\cos }^{-1}\left(1\right)-{\tan }^{-1}\left(1\right)=\frac{\pi }{2}-\frac{\pi }{4}\Rightarrow \frac{\pi }{4}$